Exam Questions KJ 3022

Audun F. Bueneaudun.f.buene@gmail.com

1. desember 2013

These questions are answered by many contributors. If you find any errors, pleaselet me know. Hopefully these short answers will help you with the theoreticalquestions.

1) What is the macroscopic magnetization? Draw the scheme.It is the sum of all the z-components of all the nuclear magnetic moments in asample (M0). M0 exists because Nα>Nβ.

Figur 1: Macroscopic magnetization shown as a sum of vectors.

2) What is the connection between a nuclear angular momen-tum and a nuclear spin quantum number? Write down theequation.The nuclear angular momentum is given by the nuclear spin quantum number.

P =√

I(I +1)ħ

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P = Nuclear angular momentI = Spin quantum numberħ = h/2, h = Planck’s constant

3) What is the connection between a nuclear angular momen-tum and a magnetic moment? Write down the equation.The magnetic moment is proportional to the angular momentum, by the proportio-nality factor γ. (The magnetogyric ratio is a constant that varies for each nuclide(isotope of a given element). The value of γ determines the sensitivity of the nucli-de in NMR, where a higher value means a higher sensitivity.)

µ= γP

µ = Magnetic momentγ = Magnetogyric ratioP = Nuclear angular moment

4) What is the Larmor frequency?The larmor frequency is the frequency of the precession of nuclear dipoles aroundthe z-axis.

Figur 2: ω indicates the larmour frequency.

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5) Does the Larmor frequency depend on the static externalmagnetic field?The Larmor frequency νL is proportional to the magnetic flux density B0.νL = | γ2π |B0νL = Larmor frequencyγ = Magnetogyric ratioB0 = Magnetic flux density

6) Which values can a nuclear spin quantum number have?0, 1/2, 1, 3/2, 2, 5/2 . . .

7) The 147 N nucleus has the nuclear spin quantum number:

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8) How many values can the magnetic quantum number have?The magnetic quantum number m can have the following values: m = I, I - 1, .... ,-IA sum of (2I+1) different values of m.

9) Write down possible values m can have for the proton.Draw the scheme.A proton has spin 1/2, the sum of different values of m is therefore (2 · 1/2 + 1) =2, which gives m = I and -I, ergo +1/2 and -1/2.

10) Write down possible values m can have for a nucleus withI=1. Draw the scheme.+1, 0 and -1.

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Figur 3: Energy level diagrams for I = 1/2 and I = 1.

11) What is the directional quantization?Directional quantization is the (number of possible) orientations of the angularmomentum and the magnetic moment in the magnetic field.

12) Explain α- and β-spin states of spin-1/2 nuclei. Draw thescheme.Due to directional quantization only certain angles of the precessing nuclear dipo-le are allowed. α-spin is with positive magnetic quantum number (m= + 1/2) andβ-spin with negative (m = -1/2).

13) Write down the equation for energy difference betweentwo adjacent energy levels of magnetic dipole in an externalmagnetic field.∆E = γħB0

Figur 4: Energy level difference due to B0 illustrated.

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14) What is the saturation?Saturation is when the N-populations are equal (Nα=Nβ). The population is grea-ter in the lower energy level, and this leads to the transition from the lower to thehigher level to become the dominant process and is therefore observed signal. Theprocess of moving up in energy require absorption of energy, while the transitiondown requires emission. Under saturation the absorption and emission are equalin intensity and cancel each other out, and thus no signal is observed. Used forbroadband decoupling, and selective removal of signals.

15) Which transitions are allowed by the spectroscopic quan-tum mechanical rules?Both transitions from lower to upper energy level due to energy absorption andfrom upper to lower due to emission of energy are allowed. The magnetic quantumnumber m can only change by a value of 1.

16) What is the pulse?The pulse is the radiofrequency pulse that excites all nuclei of one species si-multaneously. It contains a band of frequencies symmetrical about the center fre-quency ν1, with a fixed duration τP (usually several µs).

17) What happens with microscopic magnetizations after the90◦ pulse?A small fraction of the microscopic magnetizations (nuclear angular momentaP) are now processing around the z-axis in phase coherence. This results in amacroscopic magnetisation in the y’ direction. On top of this the number of angu-lar momenta in alpha state is equal to that in the beta state, which results in thedisappearance of the macroscopic magnetization in the z direction. (See Figure1-12, page 12) Phase coherence; see Figure 1-14, page 14 in the textbook.

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Figur 5: Macroscopic magnetization after a 90◦y′ pulse.

18) What is the relaxation?Immediately after a pulse the spin system will start to revert to its equilibriumstate. This is relaxation. Relaxation occurs as transverse (spin-spin) and longi-tudinal (spin-lattice). The dipoles will dephase, reducing My until zero, restoringtheir random distribution in precession around the z-axis. Nα and Nβ return totheir equilibrium state, gradually increasing Mz. (See textbook Figure 1-14, page14)

19) What is the purpose of the Fourier transformation?Its purpose is to transform the time domain spectrum, the FID, recorded duringanalysis into readable data in the frequency domain.

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Figur 6: A few FID with their corresponding fourier transformed spectras in the fre-quency domain.

20) What is FID (free induction decay)?It is the decay of the transverse magnetization, which is detected by the NMRapparatus. It is often the detected in the y’-direction. (Textbook Fig 1.14, 1.15,1.16)

21) Write down the expression showing the dependence ofthe S/N ratio on the number of scan.The signal/noise-ratio is proportional to the number of scans performed.S:N =

pNS

S:N = Signal to noise ratioNS = Number of scans

22) What is the acquisition time?The time used to obtain the FID per scan (time of relaxation between each pulse).

23) Give a list of main parts of a pulse NMR instrument.Magnet: Cryomagnets capable of a frequency up to 900 MHz in protons. (21,14Tesla, Hz is relative to the nuclei in question)Probe-head: Where the sample is insertedTransmitter: Radiofrequency generator and frequency synthesizer. Also capableof pulse production.Receiver: Radiofrequency voltage proportional to the transverse magnetization

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My′ is induced in the receiver coil.Computer: Digitizes the information and (Fourier) transforms it to readable data.

24) What is the chemical shift?The effect of the magnetic field (Be f f ) on a specific molecule is always less thanthe applied field (B0). This is due to shielding effect (σ) on the nuclei which is ana-lyzed. As nuclei are shielded differently due to their molecular environment theywill give separate resonance signals in the spectrum, creating a specific chemicalshift for a specific nuclei. The different Larmour frequencies of the nuclei resultin different chemical shifts.

25) What is the scalar coupling?It is the interval between the two signals of a doublet. It’s the splitting of a signaldue to interactions through the chemical bonds. J coupling.

26) What is the zero-order spectrum?It is a spectrum containing only singlets.

27) What is the first-order spectrum?It is a spectrum containing multiplets which follow the multiplet rules, meaningthat the shape of the multiplets are of consequent size and shape (1:1, 1:2:1,1:3:3:1 etc. . . Pascals’ triangle.) and where the frequency interval between nucleiis large compared to the coupling constant (∆ ν>> J).

28) What is the spectrum of higher order?It is a spectrum containing multiplets which does not follow the multiplet rules.The condition ∆ ν>> J is not met, altering the intensity ratios within the multip-lets (no longer following Pascals’ model) causing additional lines to appear in thespectra and making it more complex.

29) What are the 13C satellite spectra?These are signals observed due to coupling between proton and 13C . Observedas very small doublets evenly spaced around the proton signal. In proton NMR13C satellites are usually 1,1 % of the intensity of the main signal, due to the lowabundance of 13C .

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30) What are integrals?The area of each signal. In 1H NMR the integrals corresponds with the relativeamount of nuclei with the same chemical shift. Integrals from 13C NMR is gene-rally not accurate due to NOE.

31) Which are the reasons for integration of 13C spectra beingnot very common practice?Due to the low sensitivity of 13C spectra each shift is made of few data points.Which makes each shift different and the integration unreliable simply becauseof the resolution of the spectras. The system usually does not have time to resto-re equilibrium between two pulses. Because the range of carbon shifts is so large,there is a considerable frequency difference between the RF pulse and the larmourfrequency of a majority of the carbons in the sample, leading to lower intensities.Finally the NOE distorts everything when you use broadband decoupling of pro-tons. (See the four points in 1.6.3.2, page 35 for further details.)

32) The schemes below (Figures a, b and c) show pulse se-quences of three double-resonance NMR experiments. Foreach of them answer:

(a) (b) (c)

a) Which experiment is in the matter?a = Inverse gated decoupling, b = Broadband 1H decoupling, c = Gated decoupling

b) What is achieved by this experiment?a: All proton coupling is removed but no NOE is built up. (no amplification/distortion)b: All proton coupling is removed. NOE amplification allows for faster experiment.c: C-H coupling is observed while some NOE amplification is retained.

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33) Which types of relaxation you know?- Spin-lattice/Longitudinal relaxation (T1): Relaxation is applied in field direction.- Spin-spin/Transverse relaxation (T2): Relaxation perpendicular to field direction.

34) Through which mechanisms does relaxation occur?T1:Dipole-dipoleSpin-rotationChemical shift anisotropyScalar couplingElectric quadrupolarInteractions with unpaired e- in paramagnetic compoundsT2:Magnetic field inhomogeneity

35) What is the correlation time, τC?Corresponds roughly to the interval between two successive reorientations or po-sitional changes of the molecule.

36) Describe in 5 steps the inversion-recovery experiment forquick estimation of T1 relaxation time constant.A series of 13C spectra with pulse sequence 180◦

x′ −τ−90◦x′ - FID while simultane-

ously 1H BB decoupling.

The 180◦x′ pulse turns the macroscopic magnetization in the -z direction after

which it starts to relax and turn towards the -y’ axis. The amount of relaxation iscontrolled by τ, the time between the two pulses. Low τ means a short relaxationtime which means that the magnetization vector will not have time to relax pastthe -y’ axis, meaning a negative signal. Conversely, a long τ means long relaxationand a positive signal. At one time, τ = τzero, there is no signal and T1= ln(2)/τzero.

37) What is the effect of the pulse field gradient on transversemagnetization?The pulsed field gradient will set up a linear magnetic field gradient along thesample tube, meaning that every nuclei will feel a slightly different magnetic field.

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This affects the frequency with which they precess around the z-axis in the rota-ting frame of reference. After a time of about 1 ms, all the nuclei will have spreadout, completely cancelling the transverse magnetization.

It is possible to recover the transverse magnetization by inducing another fieldgradient in the opposite direction for the same period as the first.

38) What is achieved by the SPI (selective population inver-sion) experiment?In the SPI experiment, one selectively transfers the magnetization from one nucleusto another, typically 1H to 13C in order to improve the sensitivity of the carbonnucleus. In a 1H spectrum 13C satellites are often seen, and these are the tran-sitions targeted in the SPI experiment. The transfer of magnetization is achievedby a selective 180 degree pulse for the sensitive nucleus (1H). This inverts the po-pulations of one of the 13C satellites, which transfers the magnetization from theproton to the carbon, leading to a much larger signal in 13C NMR.

39) What is the difference between the APT (attached protontest) and the DEPT experiments?In the APT CH3 and CH gives positive signals, while the CH2 and quaternarycarbon gives negative signals. The DEPT experiment can be divided into threesubspectra, one for primary (CH3), one for secondary (CH2) and one for tertia-ry (CH) carbons by using different pulse-field gradients. Quaternary carbons arenot visible, but can be found by comparing the three subspectra with the 1H-BBdecoupled 13C-spectra and identifying the peaks missing.

40) The figure below shows two magnetization vectors in x’,y’plane. Draw on the same picture the effect of 90◦

y′ pulse onthem (draw their new positions).Textbook Fig 8.3, page 191. The vectors rotate clockwise around the axis of thepulse. A 90◦

y′ pulse will rotate both vectors 90◦ around the y’-axis, leaving them inthe y’,z-plane.

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Figur 7: Vectors A and B are flipped 90◦ into the y’, z plane.

41) Draw the basic spin-echo pulse sequence and show itseffect on an AX spin system by using Bloch’s vector model.90◦

x′ −τ−180◦x′ −τ−echo 1−τ−180◦

x′ −τ−echo 2−τ−180◦x′ −echo 3...

Figur 8: Spin-echo pulse sequence.

Vector model: Textbook Fig 7.4 C (a-h).

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42) What is the purpose of the polarization transfer experi-ments?It enhances the signal for insensitive nuclei and nuclei with negative magneto-gyric ratios. This is achieved by transferring the polarization from an insensitivenuclei.

43) Draw the scheme of energy levels of the heteronucleartwo spin system AX.Fig. 8.8, page 200.

44) In the reverse INEPT experiment, it is necessary to applyphase cycling. Why?Reverse INEPT is used to look for weak 13C satellites and the main signal will stillbe clearly visible. Phase cycling is used to suppress the main signal, leaving onlythe satellites. Furthermore, the two signals from reverse INEPT are of differentintensity to begin with. This is caused by the different lengths of the MH vectors.Phase cycling equalizes the two signals.

45) What information is available from the TOCSY experi-ment?Coupling between protons and other nuclei over long distance, not only neighbo-ring nuclei. Will show coupling if there is a complete chain of coupled protons.Used typically to decide which protons are situated on different monomers in apolysaccharide chain.

Figur 9: TOCSY correlations.

46) What is shimming?Adjustments for optimizing the field homogeneity.

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47) Why is it necessary to use deuterated solvents for NMRexperiments?Because the proton signal from non-deuterated solvents will easily overshadowany signals from your sample. In addition, it’s necessary to have nucleus for thelocking of the signal. In polysaccharides the deuturated solvent can exchange theprotons of the OH-groups with deuterium, making the spectras a lot easier tointerpret!

48) Which tasks an NMR probe has to perform?Transmitting and receiving signals, decoupling, locking, generating field gradi-ents and amplification of signals.

49) Presence of paramagnetic impurities is mostly unwantedfor NMR experiments. Why, what is a typical paramagneticimpurity and how it is usually removed from the sample?Paramagnetic impurities shorten relaxation time and thereby increase band broade-ning. Nuclei interacting with impurities may also have its shift moved. Dissolvedoxygen is a typical impurity and is removed by degassing the sample.

50) When and why are paramagnetic compounds deliberate-ly added to the sample before running an NMR experiment?It may be used to remove nuclear overhauser effect but also to change the shifts inorder to aid spectral analysis. If the molecule contain many similar, overlappingshifts which are hard to discern, addition of paramagnetic compound may chelatewith the molecule and change its shifts so that they are discernable.

51) What is locking and what is its purpose?Ensures the necessary field frequency stability. If there is no resonance condition,the lock unit applies a field correction until resonance is restored.

52) What is the phase correction?Removes the dispersion component signals.

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53) Which requirements a good NMR solvent has to satisfy?Good solubility of sample, stable, isotope-labeled (deuterated), non-reactive withsample and have low viscosity. Easily removable solvents may be preferred if thecompound is to be retrieved after analysis. Traces of water should be avoided asresidual HDO can give broad peaks. 13C-solvents should not overlap with sample.

54) On the scheme of a typical 2D experiment below write onthe dashed lines names of the phases.Preparation - Evolution - Mixing - Detection

Figur 10: Typical 2D pulse sequence.

55) Using the scheme above explain how the second dimensioncan be generated.The difference between 1D and 2D spectra is the the evolution time, t1. This timeis variable and it changes the amplitude of the signal peaks. This is similar to,but not the same as the inversion recovery experiment. The result of changing t1between t1 = 0 and t1 = 1/[J(C,H)] is an amplitude modulated signal, that is, theresulting signal varies in amplitude from short to long t1 values. This amplitudemodulated signal gives the F1 dimension in which peaks are separated by J(C,H).The second dimension, F2, is simplythe chemical shift of the peaks in question.

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56) The 2D HETCOR and HSQC experiments give practicallythe same information about the molecule. What is the maindifference between them?The same results but with considerably larger sensitivity in HSQC. HETCOR isrecorded on the carbon channel, while HSQC and HMQC are recorded on theproton channel.

57) Can direct C-H connections be seen in an HMBC spec-trum?No. They are suppressed by suitable phase cycles.

58) What information about a compound can be obtained fromthe 2D INADEQUATE experiment?It can assign which carbons are connected to each other. Enabling quick identifi-cation of which signals are which carbons.

59) If the W0 (zero-quantum relaxation process) is the domi-nant one, how that affects the NOE?It reduces the signal intensity. Page 300.

60) What can be said about correlation time for small mole-cules?Correlation time for small molecules is smaller as the molecule can move faster.

61) Does the external field B0 influence NOE?Yes, its signal amplification is dependent upon the magnetic field strength.

62) What improvement offers ROESY in comparison to NOE-SY?In NOESY the relative signs of the amplitudes of the cross peaks depend on thecorrelation time and sometimes the cross peaks may disappear. This effect is rem-oved in ROESY.

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63) How close protons must be in order for NOE correlationbetween them to be detected?At least less than 5 Å, ideally less than 3 Å.

64) Which interactions NOE is based on?A continuous saturation of a transition of one nucleus and the system will try torestore the equilibrium through spin-lattice relaxation.

65) What is the purpose of using shift reagents in NMR?If the molecule contain many similar, overlapping shifts which are hard to discern,addition of a shift reagent may chelate with the molecule and change its shifts sothat they are discernable. Similar to paramagnetic compound addition.

66) Is it possible to distinguish enantiomers by NMR?Not without further action. A chiral lanthanide shift reagent or a chiral diamag-netic reagent may be added to form diastereomers. The different diastereomericcomplexes may then be distinguished by their different resonances.

67) What is always the result of the rapid dynamic processeson proton NMR spectrum?They lead to a simplification of the spectrum.

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68) On the figure below NMR signals of two protons exchan-ging their positions in a dynamic process are shown. Howwill the increase of temperature affect those signals?

Figur 11: Due to a rapid dynamic process, the spectrum shown in B is simplified at highertemperature to spectrum A.

Increase of temperature will lead to one single shift representing both protons.Shift in the middle of the previous shifts. The signals will merge due to the rapidinversion of their positions, making them ‘equal’.

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